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题目：
Given a sorted linked list, delete all nodes that have duplicate numbers, leaving only distinct numbers from the original list.（Medium）
For example,
Given 1->2->3->3->4->4->5, return 1->2->5.
Given 1->1->1->2->3, return 2->3.
分析：
链表去重的follow-up，但是题号这个在前面，要把有重复元素，则所有该元素均删掉。
注意：
1. 使用dummy node处理头结点被删除掉的情况。
2. 只要有head-> next -> val存在时，先判断head -> next是否为空。
代码：
1 /** 2 * Definition for singly-linked list. 3 * struct ListNode { 4 * int val; 5 * ListNode *next; 6 * ListNode(int x) : val(x), next(NULL) {} 7 * }; 8 */ 9 class Solution { 10 public: 11 ListNode* deleteDuplicates(ListNode* head) { 12 ListNode dummy(0); 13 dummy.next = head; 14 head = &dummy; 15 while(head -> next != nullptr && head -> next -> next != nullptr) { 16 if (head -> next -> val == head -> next -> next -> val) { 17 int val = head -> next -> val; 18 while (head -> next != nullptr && head -> next -> val == val) { //只要取head->next，先判断 19 ListNode* temp = head -> next; 20 head -> next = head -> next -> next; 21 delete temp; 22 } 23 } 24 else { 25 head = head -> next; 26 } 27 } 28 return dummy.next; 29 } 30 };
LeetCode82 Remove Duplicates from Sorted List II
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原文地址：http://www.cnblogs.com/wangxiaobao/p/5940212.html
